Relation between diagonalizibilty of linear operator and its characteristic polynomial?

Question

In Hoffman-Kunze (Linear Algebra) there is this theorem :

The first two statements are equivalent, does this means if characteristic polynomial can be factored in the above way then T is diagonalizable?

But consider :

Characteristic Polynomial of A satisfies statement 2 but matrix A is not diagonalizable, So statement 1 and statement 2 are not equivalent. Am I missing something, Please Help.

Answer 1

First: The second statement has two clauses. It asserts that two things happen:

1. The characteristic polynomial splits; and
2. The dimension of the eigenspace of $\lambda$ equals the multiplicity of $\lambda$ as a root of the characteristic polynomial for every eigenvalue $\lambda$.

So the second statement is not just about being able to factor the characteristic polynomial.

Second: by my calculations, your matrix has characteristic polynomial (expanding along the second row) $$\begin{align*} \det(A-tI) &= \det\left(\begin{array}{ccc} 6-t & 3 & -8\\ 0 & -2-t & 0\\ 1 & 0 & -3-t \end{array}\right)\\ &= (-2-t)\det\left(\begin{array}{cc} 6-t & -8\\ 1 & -3-t \end{array}\right)\\ &= (-2-t)\Bigl( (6-t)(-3-t) + 8\Bigr)\\ &= (-2-t)(t^2 -3t - 10) = -(t+2)^2(t-5). \end{align*}$$

Let us write then $c_1=-2$, $d_1=2$, $c_2=5$, $d_2=1$, using the notation of the theorem.

It is true that it is not diagonalizable, because $W_1$, the nullspace of $A+2I$, has dimension $1$. But this means that it does not satisfy the second statement, because $\dim(W_1) = 1 \lt 2=d_1$. That is, $A$ does not satisfy the second statement in the theorem, because it has two satisfy two clauses and it only satisfies one of them.

Answer 2

$(\textrm{ii})$ In other words what it says is that $\textsf T$ is diagonalizable if and only if the dimension of each eigenspace associated with $c_i$ is equal to the multiplicity of the same eigenvalue $c_i$.

In your case, you forgot to check the dimension.